Ncert Exercises & Solutions

7.10 At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.

2SO2(g) + O2(g) 2SO3 (g)

What is Kc at this temperature?

For the given reaction,

∆ n = 2 - 3 = - 1

T = 450 K

R = 0.0831 bar L bar K^{- 1} {mol}^{- 1}

K_{p} = 2.0 \times 10^{10} {bar}^{- 1}

We know that,

K_{p} = K_{c} (RT) ∆ n

\Rightarrow 2.0 \times 10^{10} {bar}^{- 1} = K_{c} {(0.0831 l bar K^{- 1} {mol}^{- 1} \times 450 K)}^{- 1}

\Rightarrow K_{c} = \frac{2.0 \times 10^{10} {bar}^{- 1}}{{(0.0831 L bar K^{- 1} {mol}^{- 1} \times 450 K)}^{- 1}}

= (2.0 \times 10^{10} {bar}^{- 1}) (0.0831 I L bar K^{- 1} {mol}^{- 1} \times 450 K)

= 74.79 \times 10^{10} L {mol}^{- 1}

= 7.48 \times 10^{11} L {mol}^{- 1}

= 7.48 \times 10^{11} M^{- 1}